Asked 6/27
2071 Shrawan · 10 marks

Root locus of K/[s(s+6)(s²+4s+13)]

Short answer

Breakaway point is at s = −4.20. Angles of departure from the complex poles −2 ± j3 are ∓70.56°. The system is stable for 0 < K < 227.8; at K = 227.8 it oscillates at 2.79 rad/s, and for larger K it is unstable.

The open loop transfer function of a control system is given by G(s)H(s) = K/[s(s+6)(s²+4s+13)]. Sketch the root locus for 0 ≤ K ≤ ∞ and determine the breakaway point, the angle of departure from complex poles and the stability conditions.

G(s)H(s) = \frac{K}{s(s+6)(s^2+4s+13)}

Characteristic equation: $s^4 + 10s^3 + 37s^2 + 78s + K = 0$ .

1. Poles and zeros

Poles: $0,\ -6,\ -2 \pm j3$ . No zeros. Hence $n = 4$ , $m = 0$ , and 4 branches go to infinity.

2. Real-axis locus

The segment $[-6, 0]$ .

3. Asymptotes

\sigma_A = \frac{0 - 6 - 2 - 2}{4} = -2.5, \qquad \theta_A = \pm 45^\circ,\ \pm 135^\circ

4. Breakaway point

K = -(s^4 + 10s^3 + 37s^2 + 78s), \qquad \frac{dK}{ds} = -(4s^3 + 30s^2 + 74s + 78) = 0

Roots: $s = -4.20$ , $s = -1.65 \pm j1.39$ . The complex roots give complex $K$ and are rejected. Breakaway point: s = −4.20, with

K = |{-4.2}|\,|{1.8}|\,|17.64 - 16.8 + 13| = 4.2 \times 1.8 \times 13.84 = 104.6

5. Angle of departure from $-2 + j3$

From	Angle to $-2+j3$
pole $0$	$123.69^\circ$
pole $-6$	$36.87^\circ$
pole $-2-j3$	$90^\circ$

\phi_d = 180^\circ - (123.69^\circ + 36.87^\circ + 90^\circ) = -70.56^\circ

From $-2 - j3$ , the departure angle is $+70.56^\circ$ .

6. $j\omega$ -axis crossing

Row	Col 1	Col 2	Col 3
$s^4$	1	37	$K$
$s^3$	10	78
$s^2$	$(370-78)/10 = 29.2$	$K$
$s^1$	$(2277.6 - 10K)/29.2$
$s^0$	$K$

$K_{mar} = 227.76$ . Auxiliary equation: $29.2s^2 + 227.76 = 0 \Rightarrow s = \pm j2.79$ .

Sketch

                                       |
    ...                                |      ....
      ....                             |   ....
         ....                x.        .....
            ...               .........X
              ...                      |
                ...                    |
                  ..                   |
----------x........*...................x----------
                  ..                   |
                ...                    |
              ...                      |
            ...               .........X
         ....                x.        .....
      ....                             |   ....
    ...                                |      ....
                                       |
         -6      -4.2       -2         0
x pole   * breakaway (-4.2)
X jw crossing (+-j2.79, K = 227.8)

Branches from $0$ and $-6$ meet at $-4.2$ , break away vertically, and follow the ±135° asymptotes.
Branches from $-2 \pm j3$ leave at $\mp 70.6^\circ$ , cross the $j\omega$ -axis at $\pm j2.79$ , and follow the ±45° asymptotes.

Stability conditions

K	System
$0 < K < 227.76$	Stable
$K = 227.76$	Marginally stable, oscillates at 2.79 rad/s
$K > 227.76$	Unstable

Answer: Breakaway at s = −4.20. Departure angles are ∓70.56°. The system is stable for 0 < K < 227.8.

✓ Answer checked by Claude Opus 5.5 (high effort)

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